Calculation of integral $\int\exp \left(-\alpha \sin^2 \left(\frac{x}{2} \right) \right) dx$ Given $\alpha$ is a constant. How to calculate the following integral?
\begin{equation}
\int \exp \bigg(-\alpha \sin^2 \bigg(\frac{x}{2} \bigg) \bigg) dx
\end{equation}
Thanks for your answer.
 A: The indefinite integral cannot be expressed in terms of elementary functions. See Liouville's theorem and the Risch algorithm for more details. However, for $n\in$ Z we have the following identity in terms involving Bessel functions:
$$\int_0^{n\pi}e^{-\alpha\sin^2\Big(\tfrac x2\Big)}dx=\frac{n\cdot\pi}{\sqrt{e^\alpha}}\cdot\text{Bessel I}_0\bigg(\frac\alpha2\bigg)$$
A: A series solution is given via the Jacobi-Anger expansion:
$$-\alpha\, \left( \sin \left( \frac{1}{2}\,x \right)  \right) ^{2}=\frac{1}{2}\,\alpha
\, \left( \cos \left( x \right) -1 \right) $$
$${{\rm e}^{\frac{1}{2}\,\alpha\, \left( \cos \left( x \right) -1 \right) }}={
{\rm e}^{-\frac{1}{2}\,\alpha}}\,{{\rm I}_0\left(\frac{1}{2}\,\alpha\right)}+2\,{
{\rm e}^{-\frac{1}{2}\,\alpha}}\sum _{n=1}^{\infty } 
{{\rm I}_n\left(\frac{1}{2}\,\alpha\right)}\cos \left( nx \right)$$
where:
$${i}^{n}{{\rm J}_n\left(-\frac{1}{2}\,i\alpha\right)}=
{{\rm I}_n\left(\frac{1}{2}\,\alpha\right)}$$
was used to relate the Bessel function of the first kind $J$ to the modified Bessel function $I$. Integration term by term gives:
$$\int \!{{\rm e}^{\frac{1}{2}\,\alpha\, \left( \cos \left( x \right) -1
 \right) }}{dx}={{\rm e}^{-\frac{1}{2}\,\alpha}}
{{\rm I}_0\left(\frac{1}{2}\,\alpha\right)}x+2\,{{\rm e}^{-\frac{1}{2}\,\alpha}}
\sum _{n=1}^{\infty }{\frac {{{\rm I}_n\left(\frac{1}{2}\alpha\right)}
\sin \left( nx \right) }{n}}$$
For integration between zero and integer multiples of $\pi$ the sum clearly vanishes and you recover @Lucian's solution.
A: Let $u=\dfrac{x}{2}$ ,
Then $x=2u$
$dx=2~du$
$\therefore\int e^{-\alpha\sin^2\frac{x}{2}}~dx$
$=2\int e^{-\alpha\sin^2u}~du$
$=2\int\sum\limits_{n=0}^\infty\dfrac{(-1)^n\alpha^n\sin^{2n}u}{n!}du$
$=2\int\left(1+\sum\limits_{n=1}^\infty\dfrac{(-1)^n\alpha^n\sin^{2n}u}{n!}\right)du$
For $n$ is any natural number,
$\int\sin^{2n}u~du=\dfrac{(2n)!u}{4^n(n!)^2}-\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin^{2k-1}u\cos u}{4^{n-k+1}(n!)^2(2k-1)!}+C$
This result can be done by successive integration by parts.
$\therefore2\int\left(1+\sum\limits_{n=1}^\infty\dfrac{(-1)^n\alpha^n\sin^{2n}u}{n!}\right)du$
$=2u+2\sum\limits_{n=1}^\infty\dfrac{(-1)^n\alpha^n(2n)!u}{4^n(n!)^3}-2\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n\alpha^n(2n)!((k-1)!)^2\sin^{2k-1}u\cos u}{4^{n-k+1}(n!)^3(2k-1)!}+C$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n\alpha^n(2n)!2u}{4^n(n!)^3}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n\alpha^n(2n)!((k-1)!)^2\sin^{2k-1}u\cos u}{2^{2n-2k+1}(n!)^3(2k-1)!}+C$
$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n\alpha^n(2n)!x}{4^n(n!)^3}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{(-1)^n\alpha^n(2n)!((k-1)!)^2\sin^{2k-1}\dfrac{x}{2}\cos\dfrac{x}{2}}{2^{2n-2k+1}(n!)^3(2k-1)!}+C$
Or you can express in terms of the incomplete bessel K function:
$\int e^{-\alpha\sin^2\frac{x}{2}}~dx$
$=\int e^\frac{\alpha(\cos x-1)}{2}~dx$
$=e^{-\frac{\alpha}{2}}\int e^\frac{\alpha\cos x}{2}~dx$
$=e^{-\frac{\alpha}{2}}\int e^\frac{\alpha(e^{ix}+e^{-ix})}{4}~dx$
$=e^{-\frac{\alpha}{2}}\int e^{\frac{\alpha e^{-ix}}{4}+\frac{\alpha}{4e^{-ix}}}~dx$
Let $u=e^{-ix}$ ,
Then $x=i\ln u$
$dx=\dfrac{i}{u}du$
$\therefore e^{-\frac{\alpha}{2}}\int e^{\frac{\alpha e^{-ix}}{4}+\frac{\alpha}{4e^{-ix}}}~dx$
$=ie^{-\frac{\alpha}{2}}\int\dfrac{e^{\frac{\alpha u}{4}+\frac{\alpha}{4u}}}{u}du$
$=ie^{-\frac{\alpha}{2}}\int_0^u\dfrac{e^{\frac{\alpha t}{4}+\frac{\alpha}{4t}}}{t}dt+C$
$=ie^{-\frac{\alpha}{2}}\int_0^1\dfrac{e^{\frac{\alpha ut}{4}+\frac{\alpha}{4ut}}}{ut}d(ut)+C$
$=ie^{-\frac{\alpha}{2}}\int_0^1\dfrac{e^{\frac{\alpha ut}{4}+\frac{\alpha}{4ut}}}{t}dt+C$
$=ie^{-\frac{\alpha}{2}}\int_\infty^1te^{\frac{\alpha u}{4t}+\frac{\alpha t}{4u}}~d\left(\dfrac{1}{t}\right)+C$
$=ie^{-\frac{\alpha}{2}}\int_1^\infty\dfrac{e^{\frac{\alpha t}{4u}+\frac{\alpha u}{4t}}}{t}dt+C$
$=ie^{-\frac{\alpha}{2}}K_0\left(-\dfrac{\alpha}{4u},-\dfrac{\alpha u}{4}\right)+C$ (according to http://artax.karlin.mff.cuni.cz/r-help/library/DistributionUtils/html/incompleteBesselK.html)
$=ie^{-\frac{\alpha}{2}}K_0\left(-\dfrac{\alpha e^{ix}}{4},-\dfrac{\alpha e^{-ix}}{4}\right)+C$
